00 fundamentals maintenance management 2006

THE PROBLEM SOLVING PROCESS

Training course:

Fundamentals Maintenance Management

Author:

Prof. Dr. Attia H. Gomaa

Head of Industrial Eng. Department - Fayoum University

Industrial Engineering Consultant - AUC

Maintenance Engineering Consultant - EMC

[email protected]

2006

Fundamentals Maintenance Management

Author:

Dr. Attia H. Gomaa

Head of Industrial Eng. Department - Fayoum University [email protected]

Who Should Attend:

Managers, engineers, and other practitioners concerned with maintenance planning and control in government, industrial and services sectors.

Objectives:

To provide the participants with the modern concepts and techniques in maintenance planning and control.

To train the participants on how to use and apply these techniques in practice.

To enhance the participants experience by discussing some maintenance management problems and how to deal with them.

Course Outline:Level I: Traditional Maintenance Management1. Maintenance Management Overview

2. Preventive Maintenance Management

3. Maintenance Control

4. Computer Applications

5. PM Case Studies

6. Machine Failure Analysis

Level II: Advanced Maintenance Management7. Predictive Maintenance Management

8. Risk Based Inspection

9. Reliability Centered Maintenance

10. Total Productive Maintenance

11. Practical cases.

Level I

traditional maintenance management

1. Maintenance Management Overview

What is Maintenance?

BS 3811:1974Maintenance is defined as:

The work under taken in order to keep or restore a facility to an acceptable standard level.

Or

The combination of activities by which a facility is kept in, or restored to, a state in which it can perform its acceptable standard.

Maintenance Policies

To Keep

Planned Maintenance To Restore

Unplanned Maintenance

- Time Based Maintenance

- Condition Based Maintenance

- Risk Based Maintenance- Corrective Maintenance

- Run To Failure

- Emergency Maintenance

- Break down Maintenance

Preventive maintenance - Time-based PM

Pure time )calendar) based: Weekly, monthly, annually, etc.

Used (running) time based: 1000 km, 1000 RH, 3000 RH, etc.

Predictive (Condition-based) Maintenance by monitoring key equipment parameters "Off-line or On-line" Vibration analysis

Oil analysis

Wear analysis

Noise analysis

Temperature analysis

Pressure analysis

Quality analysis

Efficiency analysis, etc.

What are the main factors, which affect the selection of Maintenance Policy?

o Manufacturing maintenance recommendation

o System availability

o Safety factors

o Production process

o Operating conditions

o Information availability

o Resource availability

o Operating & maintenance cost

o Down time cost rate

o Failure and repair characteristics

What is the Maintenance?Example:

1-

How to keep or restore the facility at acceptable standard level in certain operating conditions? System/equipment description

Main parameters

Main items

Functional block diagram

Criticality

Working conditions

2-

How to prevent the failures?

Main failures:

PM:

3-

How to discover the hidden failures?

Main failures:

Policy:

4-

How to detect the early failures?

Main failures:

Policy:

5-

How to minimize the risk of failures?

Main failures:

Risk:

Policy:

According to maintenance information availability:

(1)

Complete Information

Planned PM

70 %(2)

Incomplete information

Planned CM

20%(3)

Without information

Unplanned CM

(or Emergency)

10%

Typical Work (man-hour) distribution in engineering industries

Experience:

Technical

Planning

Analysis

Decision making

Problem solving

Working conditions, etc.

Information:

Catalog

Forms / reports

Data collection

PM levels

Job plans for each PM level

Resources

Cost rates

CM work orders

Failure analysis, etc.

Tools:

Computer programs

International standards

Management tools, etc.

What is the ratio between maintenance cost

& manufacturing costs?

Maintenance costs are a major part of the total operating costs of all manufacturing or production plants.

Depending on the specific industry, maintenance costs can represent between 15% and 40% of the costs of goods produced.

For example in food related industries, the average maintenance cost represents about 15% of the cost of goods produced; while in iron and steel, pulp and paper and other heavy industries maintenance represents up to 40% of the total production costs.

US industry spends more than $200 billion dollars each year on maintenance of plant equipment and facilities,

USA Industries in 1983/ 1984: Maintenance Cost ( $ 35 * 109 Per year

Maintenance Cost: 10 25 % &

Spare parts Cost: 3 10 %

What are the main elements of Maintenance cost?

Direct cost:

Spare parts & supplies cost

Labor cost

Contract cost

Indirect cost:

Overhead cost

Down time cost

Maintenance cost = Direct cost + Overhead cost

Maintenance Costs Elements

Cost to replace or repair

Losses of output

Delayed shipment

Scrap and rework

What is Maintenance Management (MM)?

MM is a powerful systematic methodology to maximize the facility performance and to improve the maintenance resource productivity, through optimizing the maintenance policies for the critical equipment. MM - is the application of knowledge, tools and scientific techniques to identifying and analysis the maintenance activities.

MM - decision-making process to select the best maintenance policies for improving the equipment reliability to an acceptable level.

MM is the art of matching a maintenance's goals, tasks, and resources to accomplish a goal as needed.

MM is do the right things, with the right tools, and in the right way".Through:

1. Define the target and constraints,

2. Information collecting & analysis,

3. Maintenance planning,

4. Maintenance organization,

5. Motivation & direction,

6. Maintenance control,

7. Corrective actions, and

8. Learned lessons.

Maintenance Management History

How do you measure MM success?

Maintenance Planning Concept:

Before you start to maintenance plan, consider...

Who is the ultimate customer?

What are the customer needs?

How long will the maintenance project last?

Where are we now?

Where should we end-up?

What are the cost constraints?

What are the technical challenges?So, Maintenance Planning must determines what, when, where, how, and by whom something is done.

What is to be maintained?

"Description"

Why?

"Target" How?

"Method" By whom?

"Resources" When?

"Schedule"

Where?

"Location"What are the main Types of MM Plans?

1- MM management level plans:

Master plan Top management (10 -15 activity)

Action planControl management (50-100)

Detailed planOperational management (> 500)2- MM Time plans:

Long term

2 to 10 y

Risk 15 to 25%

Medium term

6m to 1 y

Risk 7 to 10%

Short term

1w to 3 mRisk 3 to 5%3- MM risk plans:

Target plan (normal or most likely)

Optimistic plan (best case)

Pessimistic plan (worst case)

4- MM Strategic Plans:

Strategic plan

Tactical plan

Operational plan

Urgent plan

5- MM Planning Level:

Overall plan

Complete information Partial plan

Incomplete information Urgent plan

Without informationWhat is the Maintenance System?

A system is a collection of components (or items) that work together to achieve a certain objective.

Sub-system: Water Pump Unit

Figure - Functional block diagram for a pump

Figure Main Components

Current PM Program:

ItemJob planFrequency

(1)

Motor

(2)

Coupling

(3)

Pump

(4)

Suction line

(5)

Discharge line

(6)

Valves

Root Cause Failure Analysis:

ItemMain FailuresRoot CauseMTBF

(1)

Motor

(2)

Coupling

(3)

Pump

(4)

Suction line

(5)

Discharge line

(6)

Valves

1) Motor:

FailurePMPrDCM

PolicyFreq.PolicyFreq.

2) Coupling:

3) Pump:

4) Suction line:

5) Discharge line:

6) Valves:

Developed PM Program:

ItemJob planFrequency

(1)

Motor

(2)

Coupling

(3)

Pump

(4)

Suction line

(5)

Discharge line

(6)

Valves

Modern Maintenance Management Systems:

There are four modern approaches:

1- Optimal system maintenance (OSM),

2- Risk Based Inspection (RBI)

3- Reliability centered maintenance (RCM), and

4- Total productive maintenance (TPM).

Maintenance management methodologies

OSMRBI & RCMTPM

Main objective Improve equipment availabilityPreserve system function & improve system availabilityImprove overall system productivity

ApproachMaintenance information analysis and

Using optimal mathematical modelingImprove the maintenance program

System reliability analysis

Failure mode effect analysis FMEA

Risk analysisSystem overall analysis

Continuous improvement techniques

PolicyApproachGoals

ReactiveRun to failure (fix-it when broke).Minimize maintenance costs for non-critical equipment.

PreventiveUse-based maintenance program.Minimize equipment breakdown.

PredictiveMaintenance decision based on equipment condition.Discover hidden failures and improve reliability for critical equipment.

ProactiveDetection of sources of failures.Minimize the risk of failures for critical systems.

GlobalIntegrated approach.Maximize the system productivity.

PolicyApproachGoals

RCFAIdentification of root causes of failures.Eliminate failures.

FMECAIdentification of criticality of failures.Improve equipment availability.

HAZOPIdentification of hazards and problems associated with operations.Improve HSE effect.

RCMDetermination of best maintenance requirements for critical systems.Preserve system function & improve reliability.

RBIDetermination of an optimum inspection plan for critical systems.Improve system HSE and availability.

PolicyApproachGoals

OSMOptimization approach for the global maintenance system.Maximize reliability measures and minimize maintenance cost rates.

TPMComprehensive productive-maintenance system.Maximize plant effectiveness and resource productivity.

Preventive Maintenance Management

Why Preventive Maintenance should be done? To Prevent FailureTo Detect Early Failure

To Discover a Hidden Failure

Rather, it is better to consider PM only when:

1- High Down time cost rate

2- High Safety level

3- Predictive M. cannot be applied

4- CM cannot be justified

What are the main targets of PM? Improving equipment availability/reliability

Increasing equipment effective life time

Increasing resource utilization

Increasing productivity

Reducing operating cost

Reducing total cost rate

Increasing profitability ratio

PM = Profit

What are the main Elements of Planned Maintenance?

1. Inventory list

2. Layout of facilities

3. Facility register

4. Maintenance program

5. Maintenance job specification

6. Maintenance schedule

7. Job orders

8. Follow up cards

9. Performance evaluation

Note : 1 to 5 Basic data, 6 Scheduling, and 7 to 9 Follow up and evaluation.Maintenance Planning Steps:

1. System criticality analysis

2. Equipment selection

3. Information collection & analysis

4. Target & constraints definitions

5. Requirements & standard levels

6. Main failures determination

7. Root cause failure analysis (RCFA)

8. Best maintenance policy

9. Maintenance policy planning

10. Work orders

11. Measure

12. Analysis

13. Action

14. Performance evaluation & KPI

15. Improvement

Maintenance Planning Steps:StepDescription

1. System criticality analysisHSE - Process Down time Cost

2. Equipment selection Critical equipment

Non-critical equipment

3. Information collection & analysisMaintenance catalog Design information Equipment history- Working conditions- PMs CMs Trouble shooting Reliability information HSE instructions. etc.

4. Target & constraints definitions Targets: Reliability, Availability, Down time, Cost, HSE level, .. etc.

Constraints: Budget, Spare parts, Tools, Manpower, Information,etc.

5. Requirements & standard levels Functional levels: Flow rate, Head, Pressure, Power, .. etc.

HSE levels

6. Main failures determinationFunctional failures - HSE failures Mechanical failures Electrical failures - .. etc.

7. Root Cause Failure Analysis Main failures, Root cause, RRC, Mechanism, Probability, MTBF, MTTR, Remedy.

8. Best maintenance policy Run To Failure (RTF)

Time-based (Preventive) PM

Condition-based (Predictive) PdM

Risk-based (Proactive) PrM

Maintenance Planning Steps:StepDescription

9. Maintenance policy planningFrequency- Levels- Alarm limits- Tools- Job plan- HSE plan- Spare parts- Duration- Manpower- .. etc.

10. Work orders W/O # - W/O type- Dates/time - Responsibility- Level - Alarm limits- Tools- Job plan- HSE plan- Spare parts- Duration- Manpower- Failure - Root cause- .. etc.

Complete Feedback.

11. MeasureRunning hours- Noise- Vibration- Temperature- Oil level- viscosity- Flow rate Head Speed - .. etc.

12. AnalysisNoise analysis- Vibration analysis Temperature analysis - Oil analysis - Flow rate analysis Head analysis Speed analysis - .. etc.

13. Action- Good condition

- Call for service (PM)

- Call for repair (planned CM)

- Breakdown (unplanned CM)

14. Performance evaluation & KPICM/PM- MTBF- MTTR- MTBM- MTTM- Reliability Availability- Maintainability- RAM- Spare parts consumption rates- .. etc.

15. Improvement Information Maintenance levels- Tools Spare parts Manpower skills Time HSE - .. etc.

Approach: FMEA - RCM - RBI- PMIS - .. etc.

What are the main Elements of Maintenance Plan?

1- Equipment name & code,

2- Equipment priority,

3- Maintenance start time,

4- Maintenance down time,

5- Maintenance level and type,

6- Maintenance job description,

7- Maintenance operations time,

8- Maintenance effort (man-hour),

9- Manpower requirements planning,

10- Spare parts and supplies requirement planning,

11- Tools requirements planning,

12- Failure analysis,

13- Maintenance cost estimation,

14- Maintenance budget, and

15- Safety instructions.

Maintenance Work Order

Work order number

Requester Section:

Plant (or department) name / code

Equipment name / code

Equipment priority

Maintenance type & level (PM / Repair / Overhaul)

Job scope & description

Responsibility

Planning Section:

Manpower types & skills

Time estimation

Spare parts

Special tools

Expected equipment down time (from xxx to xxx)

Cost estimation

Safety instructions

Responsibility

Craft Feedback:

Job scope & description

Manpower types & skills

Time estimation

Spare parts

Special tools

Actual equipment down time (from xxx to xxx)

Actual Cost

Responsibility

Coding:

Plant (or department), Equipment

Resources (Manpower, Spare parts, Special tools)

3- Maintenance Control

Total Control Indicators:

1- Work quantity control

Over estimation

Under estimation

2- Time control

Behind schedule (late)

Ahead schedule (early)

3- Cost control

Cost overrun

Cost under-run

4- Quality control

Acceptable level

Non-acceptable level

5- Inventory control

Over estimation

Under estimation

6- Resources control

Over estimation

Under estimation

7- Plant condition control (HSE, etc.)

Acceptable level

Non-acceptable level

Control Steps:

1- What to control?

2- What is the standard (target) performance?

3- What is the actual performance level?

4- Comparison between the actual & target.

5- Detection of variance

6- Identification of causes of variance

7- Corrective actions

8- Learned lessons.

Total Control Levels:

1- Review and data collection.2- Follow-up.

3- Performance evaluation.

4- Productivity analysis.

5- Corrective actions.

6- Learned lessons.

Maintenance Control Levels:

- Maintenance Follow-up

-(Actual/Plan)- Maintenance Performance Evaluation- Time Availability

- Reliability

- Mean Time Between Failures (MTBF)- Mean Time To Failures (MTTF)- Mean time to repair (MTTR)- Mean time between repairs (MTBR)- Mean Time Between Maintenance (MTBM)- Preventive Maintenance Rate (PM rate)- Resources Productivity Analysis

Productivity Dimensions

CostQualityQuantityTime

Efficiency

1- Technical Efficiency

2- Operating Efficiency

3- Production Efficiency

4- Economical Efficiency Effectiveness

= Actual output /

Planned output

Maintenance System Effectiveness:

It is related to performance.

It is the degree of accomplishment of objectives.

How well a set of results is accomplished?

Maintenance System Efficiency:

It is related to resource utilization.

It is the degree resources utilization.

How well the resources are utilized to achieve the results.

Productivity:

It is a combination of both effectiveness & efficiency.

Productivity index

= Output obtained / Input expended

= Performance achieved / Resources consumed

Total productivity = Total output / Total input

Partial productivity = Total output / One of the inputs

Measurement of MAINTENANCE EFFECTIVENESS

Equipment Losses Categories

CategoryEquipment lossesIndicator

Down-time losses

(lost availability)Equipment failures

Set-up and adjustmentsEquipment availability

Speed losses

(lost performance)Idling and minor stoppages

Reduced speed operationEquipment performance efficiency

Defect losses

(lost quality)Scrap and rework

Start-up lossesEquipment quality

Rate

Resource lossesCritical resource consumption ratesResource productivity

Cost lossesAll the previous lossesRepair cost

CM/PM cost ratio

Down time cost

Overall equipment effectiveness (OEE)

OEE = Equipment Availability Performance efficiency Quality rate

Total effective equipment productivity (TEEP)

TEEP =Utilization Availability Performance efficiency Quality rate

Net equipment effectiveness (NEE)

NEE = Uptime ratio Performance efficiency Quality rate

Mean unit between assists (MUBA):

MUBA = Total number of units produced / Number of stoppages

What is the effect of Maintenance Policy on the Equipment OEE?

Maintenance PolicyOEE

Operate to failure (RTF) 30 50 %

Good PM Program

Good bonus & incentive system60 80 %

Good PM Program based on RCM

Good bonus & incentive systemMore than 80 %

What are the main factors, which affect the Equipment OEE?

Product quality

Production continuity & rates

Shutdown frequency HSE factors

Equipment availability

Resource availability

Operating & maintenance cost

Down time cost rate

Maintenance Risk levels:

Objective Risk Levels:

Risk %0 - 55 1010 - 1515 - 25> 25

Risk level01234

DescriptionMinorLowMediumHighMajor

Acceptable Risk limits:

Long term

2 to 10 y

Risk 15 to 25%

Medium term

6m to 1 y

Risk 7 to 10%

Short term

1w to 3 mRisk 3 to 5%Maintenance Safety Levels:

LevelSeveritySafety (people)

0NoDoes not apply

1Very low

(Slight)Slight injury

Simple first aid

2Low

(Not Serious)Minor injury

No lost time

No Hospitalize

First aid

3Medium

(Serious)Major injury

Lost time

Hospitalize

Temporary disability

4High

(Very Serious)Fatal injury

Hospitalize

Permanent disability

5Very High

(Catastrophic)Multiple fatalities

Maintenance Performance Evaluation

What are our measures?

What are the units?

What is the time frame? What data is required?

What data is available?

Quality of data

Linking data to measuresHow to measure the performance of PM program?

Four major factors that should control the extent of a PM program:

1- The cost of PM program (PM & repairs costs).

2- Equipment reliability & utilization.

3- HSE (Health, Safety and Environment) level

4- Down time cost.

Availability = A = x 100%

Percentage of downtime = Id = 100% - AMean time between failures = MTBF =

Mean time to repair MTTR =

Where,S = Scheduled production time

d = Downtime

f = Number of failures.

df = Downtime delays from failures.

Example:

Scheduled production time = 31 day

Downtime = 6 day

Number of failures = 3 failure/month

A = x 100% = 80.6 %

Id = 100 - 80.6 = 19.4%

MTBF = = 8.33 days

MTTR= = 2 days

Maintenance Administration Indicators (%):

1- Overtime hours per month

2- Worker activity level

3- Worker productivity

4- Worker utilization

5- Scheduled hours6- Preventive & predictiveMaintenance Effectiveness Indicators (%):

1- Overall effectiveness

2- Gross operating hours

3- Number of failures

4- Breakdown downtime

5- Emergency man-hours

6- Predictive & preventive

Maintenance Cost Indicators (%):

1- Maintenance cost

2- Maintenance cost/unit

3- Maintenance manpower cost

4- Subcontracted cost

5- Cost of maintenance-hour

6- Supervision cost

7- Preventive maintenance cost

8- Cost of spare partsMain Indicators Calculations:

Overtime hours per month = % =

x 100

Worker activity level = % =

x 100

Worker productivity per month = % = x 100

Worker utilization = % =

x 100Scheduled hours versus hours worked = %

= x 100

Preventive and predictive maintenance conducted as scheduled = %=

x 100

Predictive and preventive maintenance coverage% =

x 100

Overall equipment effectiveness (OEE) = A x S x Q

A = Availability indicator

S = Speed indicator

Q = Quality indicator

Availability = A =

Speed = S =

Quality = Q =

Percentage of gross operating hours % =

x 100

Number of failures in the system (NFS) =

Equipment downtime caused by breakdown % =

x 100

Emergency man-hours % =

x 100

Emergency and all other unscheduled man-hours % =

Evaluation of predictive and preventive maintenance % =

x 100

Cost of maintenance to added value of production % =

x 100

Maintenance cost per unit of production = Cost per unit

=

Manpower component in the maintenance cost % =

x 100

Cost of subcontracted maintenance =% =

x 100

Ratio of labor cost to material cost of maintenance =

Cost of maintenance-hour = $ =

Supervision cost as a percentage of total maintenance cost %=

x 100

Progress in cost reduction effects = Index =

Preventive maintenance (PM) cost as related to breakdown maintenance

% = x 100

Inventory turnover rate per year =

Rate =

Cost of spare parts and material to maintenance cost

% = x 100

Ratio of stock value to production equipment value =

4- Computerized CMMS

More than 100 Ready-Made Packages

Most common CMMS: EMPAC

www.plant-maintenance.com

FMMS

www.kdr.com.au

GPS5

www.gps5.com

IMAINT

www.dpsi.com

IMPACT-XP

www.impactxp.com

IMPOWER

www.impower.co.uk

MAINPAC

www.mainpac.com.au MAINPLAN

www.mainplan.com

MAXIMO

www.maximo.com

MP2

www.datastream.net

OEE MANAGERwww.zerofailures.co.uk

OEE SYSTEMSwww.oeesystems.com

OEE TOOLKITwww.oeetoolkit.com

OEE-IMPACTwww.oeeimpact.com

PEMAC

www.pemac.org

PERFORM OEE www.ssw.ie/performoee.asp

RAMS

www.reliability.com.au

RCM Turbo

www.strategic.com

REAL-TPI

www.abb.com

SAP-RLINK

www.osisoft.com

TPM Software

www.tpmsoftware.com

Most MMIS systems can usually:

1. Track components,

2. Provide logistic support (e.g., spares inventory),

3. Store maintenance history,

4. Alarm predetermined maintenance activities,

5. Produce management reports.

A small number of these systems are able to:

6. Analyse maintenance history, and

7. Determine optimal policies for components and sub-systems.For a complex system, MMIS will also have to:

8. Incorporate expert opinion in a knowledge base,

9. Incorporate subjective data from experts,

10. Combine maintenance activities into schedules,11. Update schedules with occurrence of events such as failures etc,

12. Plan resources, and

13. Measure the effectiveness of maintenance activities.

This requires a more quantitative and scientific approach to maintenance management.What is the effect of the Good Computerized Maintenance Package?

1- Increase labor utilization by 5 25 %

2- Increase equipment utilization by 5-15%

3- Decrease spare parts inventory by 10-20%

4- Decrease down time cost by 5-15%

CMMS Block diagram:InputsToolOutputs

1- Reference data

2- Equipment list

3- Equipment priority

4- PM information

5- Resource list

6- Working conditions7- CM information8- Cost rates

9- Other data

10-Actual performanceExcel1- Maintenance labor force.2- Average system availability.

3- Annual downtime cost losses.

4- Annual maintenance cost.

5- Annual PM plan.

6- Maintenance resources

7- Monthly PM plans.

8- Maintenance work order9- Other reports

10- Maintenance Control

CMMS main Steps:

5- PM Case StudiesCase (1):

How to construct the coding & criticality systems:

Equipment CodingLocationEquipment TypeEquipment Tag #

123478

Propose a coding system and priority rules for the following equipment:

Plant SystemsLocationEquipment TypeNumber of Machines

Productive systemsMachining shopTurning

Milling

Drilling

Grinding

Press4

2

2

2

1

Foundry shopInduction furnaces

Molding machines2

5

Welding shopArc Welding1

Supportive systemsMaterial handlingFork lift

4

Air roomCompressor2

Water roomPump 50 HP

Pump 100 HP2

2

Power roomDiesel generator2

Equipment Coding Structure:

LocationEquipment TypeEquipment Tag #

123478

LocationEquipment Type

01 Machining shop01 Turning

02 Milling

03 Drilling

04 Grinding

05 Press

02 Foundry shop10 Induction furnaces

11 Molding machines

03 Welding shop20 Arc Welding

04 Material handling30 Fork lift

05 Air room40 Compressor

06 Water room51 Pump 50 HP

52 Pump 100 HP

07 Power room06 Diesel generator

Example: 010202

010202

Machining shopMilling# 2

Example: 065201

065201

Water roomPump 100 HP# 1

EIGHT LEVEL DECOMPOSITION:

LevelCharacterization

0System

1Sub-System

2Major Assembly

3Assembly

4Sub-Assembly

5Component

6Part

7Material

Equipment PriorityFailure effect:

- Effect on HSE

- Effect on Production

- Effect on Cost

Failure Probability:

- Failure Frequency

Example:

Factors%Levels

1- Production 30V- Very Important

I- Important

N- Normal

2- HSE30V- Very Important

I- Important

N- Normal

3- Stand by15WO- Without

WS- With Standby

4- Value5H- High Value

M- Medium

L- Low

Priority LevelDescription

AGroup A: Equipment with 100% duty factor, whose failure involves production losses and potential safety hazards.

BGroup B: Equipment with a ratio duty factor, i.e., having some standby, whose failure involves production losses and potential safety hazards.

CGroup C: Equipment with standby, whose failure involves either production losses or potential safety hazards.

DGroup D: Equipment with standby, whose failure involves neither production losses nor safety hazards.

Equipment PrioritiesLocationEquipment TypePriority Level

Machining shopTurning

Milling

Drilling

Grinding

PressB

B

B

B

D

Foundry shopInduction furnaces

Molding machinesA

B

Welding shopArc WeldingA

Material handlingFork liftC

Air roomCompressorC

Water roomPump 50 HP

Pump 100 HPC

C

Power roomDiesel generatorA

Case (2):

How to select the best maintenance policy?

Number of Engine 2000

Capital maintenance policy for engine is as follows:

Four Policies:

Replacement after first failure (after 36 month)

Repair (010) after first failure & Replacement after second failure (after 30 month)

Repair (020) after second failure & Replacement after third failure (after 24 month)

Repair (030) after third failure & Replacement after fourth failure (after 15 month)Cost rate:

Replacement $ 10,000& Repair $ 3,500

Required:

Select the best maintenance policy

Estimate the annual budget for the best policy

Target maintenance plan

Case (3):The yearly maintenance information for ten gas generators (GG) in a site are as follows:

1- Working conditions for each GG:

Average working hours 7000 hour/year2- PM Levels for each GG:

Check oil level every 150 R.H. (about 2 liter) Change oil every 750 R.H. (about 20 liter) Change oil filter every 1500 R.H.3- CM for each GG:

1. Average oil quantity is 100 liter/year/G.G.

4- Cost rates:

2. Oil cost 5 $/liter

3. Filter cost 50 $/unitRequired:

1. Annual materials (oil and filters) requirements Planning.

2. Annual materials cost

3. Annual PM plans

4. Materials profile (histogram)

5. Maintenance work order for each PM level Case (4):The yearly PM programs information for six similar gas turbines in a power station are as follows:

1- PM information:

Maintenance levels per gas turbine

PM TypeFrequencyDurationNo. of

WorkersSpare parts Cost

$1000

Y Level 1Yearly15 days2010

S Level 26 Monthly10 days208

3M Level 33 Monthly5 days155

M Level 4Monthly2 days102

2- Working conditions:

Gas turbine operating conditions: 24 hour/day

Workers operating conditions: 300 day/year & 8 hour/day

3- CM information:

Average effort of CM = 380 man-day per gas turbine

Average annual spare parts CM = $ 12000 per gas turbine

Average CM downtime = 15 days/year per gas turbine

Average downtime cost rate = $ 1000 per day

4- Cost rates:

Average labor cost rate = $ 10 per man-day

Overhead cost = 25 % direct cost (spare parts & labor)

Required:

1) The size of maintenance labor force.2) Average system availability.

3) Annual downtime cost losses.

4) Annual maintenance cost.

5) Annual PM plan.

6) Maintenance resource profiles.

7) Monthly PM plans.

8) Maintenance work orderThe size of maintenance labor forcePM TypeAnnual

FrequencyDuration

(day)No. of WorkerMan-day

per PM type

Y11520300 * 1= 300

S11020200 * 1 = 200

3M251575 * 2 = 150

M821020 * 8 = 160

Annual PM man-day per gas turbine810

Total PM annual man-day Required810 * 6 = 4860

The size of PM labor force = 4860/300 =16.2 = 17 workers

The size of CM labor force = 380 * 6 / 300 = 8 workers

Total labor force = 17 + 8 = 25 workers

Crew check is ok (25 more than 20).

The average down time per year

PM TypeAnnual

FrequencyDuration

(day)PM Downtime

(day)

Y11515 * 1= 15

S11010 * 1 = 10

3M255 * 2 = 10

M822 * 8 = 16

PM downtime per gas turbine51

Average down time = 51 + 15 = 66 day/year per gas turbine

Annual downtime cost losses = 66 * 6 * 1000 = $ 396000

Average equipment availability =

Active operating time / Total time

= (364 66) / 364 = 82 %

System Reliability:

Series or chain structure: Rs = R1 * R2 * R3 * etc.

Parallel structure: Rs = 1 (1-R1)* (1-R2)* (1-R3) * .etc.

System time availability =

Parallel structure: As = 1 (1-A1)**6

= 1 (1-0.82)**6

= 1 (0.18)**6 = 99%

Annual maintenance costPM TypeAnnual

FrequencyCost

$1000Spare parts PM Cost $1000

Y11010 * 1= 10

S188 * 1 = 8

3M255 * 2 = 10

M822 * 8 = 16

Annual spare parts PM per gas turbine =44

Total annual spare parts PM cost =44 * 6 = 264

The average annual spare parts CM cost =

$ 12000 * 6 = $ 72,000

Annual spare parts maintenance cost =

264000 + 72000 = $ 336,000

Annual labor cost =

25 workers * 300 day/year * $ 10 per man-day= $ 75,000

Annual direct maintenance cost = $ 336000 + $ 75000

= $ 411000

Overhead cost = 25 % direct cost

Annual maintenance cost = $ 411000 * 1.25 = $ 513750

Annual maintenance cost = $ 513750

Annual downtime cost losses = $ 396000

Basic Annual PM Plan

Eq.

codeMonth #

123456789101112

G01YMM3MMMSMM3MMM

G02MMYMM3MMMSMM3M

G03M3MMMYMM3MMMSM

G04SMM3MMMYMM3MMM

G05MMSMM3MMMYMM3M

G06M3MMMSMM3MMMYM

Resource analysis:

Man-day580230580230580230580230580230580230

Day/

month242424242424242424242424

Workers241024102410241024102410

SP cost261826182618261826182618

DT331833183318331833183318

Y= 300S= 2003M= 75 M= 20 man-day

Y= 10S= 8

3M= 5 M= 2 $1000

Y= 15S= 103M= 5 M= 2 day

Target Annual PM Plan # 1

Eq.

codeMonth #

123456789101112

G01YMM3MMMSMM3MMM

G02MMYMM3MMMSMM3M

G03M3MMMYMM3MMMSM

G04MSMM3MMMYMM3MM

G053MMMSMM3MMMYMM

G06MM3MMMSMM3MMMY

Resource analysis:

Man-day455355455355455355355455355455355455

Workers191519151915151915191519

SP cost232123212321212321232123

DT282328232823232823282328

Y= 300S= 2003M= 75 M= 20 man-day

Y= 10S= 8

3M= 5 M= 2 $1000

Y= 15S= 103M= 5 M= 2 day

Target Annual PM Plan # 2

Eq.

codeMonth #

123456789101112

G01YMM3MMMMSM3MMM

G02MMYMM3MMMMSM3M

G03M3MMMYMM3MMMMS

G04MSM3MMMYMM3MMM

G05MMMSM3MMMYMM3M

G06M3MMMMSM3MMMYM

Resource analysis:

Man-day400410400410400410400410400410400410

Workers171717171717171717171717

SP cost202420242024202420242024

DT252625262526252625262526

Y= 300S= 2003M= 75 M= 20 man-day

Y= 10S= 8

3M= 5 M= 2 $1000

Y= 15S= 103M= 5 M= 2 day

Monthly Maintenance Plan: Month # 1

DayG01G02G03G04G05G06PM worker

1. Y20

2. Y20

3. Y20

4. Y20

5. Y20

6. Y20

7. Y20

8. Y20

9. Y20

10. Y20

11. Y20

12. Y20

13. Y20

14. Y20

15. Y20

16. SB-

17. M10

18. M10

19. SB-

20. M10

21. M10

22. SB-

23. M10

24. M10

25. SB-

26. M10

27. M10

28. SB-

29. M10

30. M10

31. SB-

Maintenance Work Order

010120

Requester Section:

Power Station PS03 - Gas Turbine G01 - Priority: A

Maintenance type/level: Annual PM1- Check .

2- Clean ..

3- Replace ..

4- Adjust

5- Repair ..

Eng. Attia Gomaa

Planning Section:

Labor: 4 Mech. 2 Helper 5 days

5 Elec. 4 Helper 10 days

Spare parts: 2 valve xx1, 4 air filter yy3, .. etc.

Special tools: xxx, yyyy, etc,

Expected down time (from 01/01 to 15/01/2004)

Cost estimation ($ 10,000)

Safety instructions:

- Check Eng. Aly Ahmed

Craft Feedback:

1- Check .

2- Clean ..

3- Replace ..

4- Adjust

5- Repair ..

Labor: 3 Mech. 2 Helper 5 days

6 Elec. 3 Helper 11 days

1 Vib. 1 Helper 2 days

Spare parts: 2 valve xx1, 4 air filter yy3, .. etc.

Special tools: Vibrometer, etc,

Down time (01/01 to 17/01/2004) Actual Cost ($ 12,000)

Eng. Omer AlyCoding:

Case (5):

The yearly maintenance information for three generators in a site are as follows:

1- Working conditions:

Two gas generators (GG01 and GG02), one operating and the other standby

Diesel generator for emergency

Site operating hours 24/day * 365 day

2- PM Levels (Catalog information):

Check oil level every 150 R.H. (about 2 liter)

Change oil every 750 R.H. (about 20 liter)

Change oil filter every 1500 R.H.

Check cooling level every 150 R.H.

Clean/ drain cooling system every 1500 R.H.

Check and clean batteries every 1500 R.H.

Lubricate bearing every 750 R.H. (about 1 liter)

Change bearing every 3000 R.H.

Replace thermostat every 3000 R.H.

3- CM for each GG:

Average oil quantity is 100 liter/year/G.G.

4- Cost rates:

Oil cost 3 $/liter

Filter cost 10 $/unit

Bearing oil cost 5 $/liter

Bearing cost 30 $/each

Thermostat cost 30 $/each

Required:

1. Maintenance work order for each PM level 2. Annual materials requirements Planning & materials cost

3. Annual PM plans

4. Cost & materials profiles (histogram)

Case (6):Maintenance spare parts cost ($):Year

1999Year

2000Year

2001Year

2002Exp.

2003Forecasting limits

2003

1450130012001000??

X12345

Y1450130012001000?

XY1450260036004000

n = 4

Sum X = 10

Sum X2 = 30

Sum Y = 4950

Sum XY = 11650

Sum Y = n . a + b Sum X , Sum XY = a Sum X + b Sum X2

4950 = 4 a + 10 b

11650 = 10 a + 30 b

14850 = 12 a + 30 b

a = 1600

b = - 145

Y = 1600 145 X

Y5 = 1600 145 (5) = 875

X12345

A1450130012001000-

F1445131011651020875

(A-F)5103520

(A-F)2251001225400

MSE = 1750 / (4 -1) = 583

S = 24

Z = 2

CLs = 0 Z S = 0 48Case (7):

Uncertain spare parts cost

Spare parts cost

$ 100,000Probability

%

920

1050

1120

127

133

Estimate the spare parts budget based on the following:

1- Average method ($ 1,100,000)

2- Probability method ($ 1,023,000)

3- PERT method ($ 1,033,000)

Solution

Maintenance shutdown Planning

Using CPM

Case (8): The monthly PM programs information for a machining shop are as follows:

Machine CodeT01D01M01T02M02

Machine DescriptionTurningDrillingMillingTurningMilling

Predecessors---T01M01

Duration (day)856820

Worker/day58755

Spare Parts cost $ 1000543612

Maximum worker limit is 12 worker/dayRequired:

1. Monthly maintenance plan.

2. Calculate the monthly spare parts cost.

3. Construct the monthly spare parts cost profile.

Case (9): Monthly Maintenance Plan for Wire Production Line

Project Name : MMPW

Project start: 1 Jan. 2004Planning unit : Day

6 DAYS /WEEK

1- Activity List

ActivityIDDuration (day)PredecessorsRelations

(SS, FS, FF, and SF)

1PreparationPRP2--

2Mech. maintenance # 01MM17PRP-

3Elec. maintenance # 01EM19MM1SS 3





8SetupSTP1EM1

EM2

EM3-

2- Resource List

Resource

CodeResources descriptionUnitLimits/dayPrice

LE/unit

Norm.Max.

L01Mechanical workermd3640

L02Electrical worker md4860

SPSSpare parts & suppliescost--1000

3- Resource Allocation

ActivityIDResource

L01/

dayL02/

daySPS

(Total)

1PreparationPRP211

2Mech. maintenance # 01MM14-3

3Elec. maintenance # 01EM1-54





8SetupSTP2

21

4- Base Calendar (Working periods)

SaturdaySundayMondayTuesdayWednesdayThursdayFriday

XXXXXX

1/01/04

Holidays:20 to 21 Jan. 2004

Required:

1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?

3. Construct the corresponding smoothed worker loading?4. Construct the corresponding worker leveling?

5. Construct the target action plan?.

6. Construct the cost profile & S-curve?7. Construct the target master plan?Case (10): Annual Maintenance Plan for AUC-IT Labs.

Project Name : AMIT

Project start: 1 Jan. 2004Planning unit : Day

6 DAYS /WEEK

1- Activity List

ActivityIDDuration (day)PredecessorsRelations

1PreparationPRP1--

2Server maintenanceSRM3PRP-

3Hardware maintenance Lab #01HM14SRM-

4Software maintenance Lab #01SM15HM1


6Software maintenance Lab #02SM24HM2SS 1


8Software maintenance Lab #03SM34HM3SS 1

9SetupSTP1SM1

SM2

SM3-

2- Resource List

ResourceCodeResources descriptionUnitLimits/dayPrice

LE/unit

Norm.Max.

L01Hardware Engineermd36120

L02Software Engineermd48100

SPSSpare parts & suppliescost--1000

3- Resource Allocation

ActivityIDResource

L01/

dayL02/

daySPS

(Total)

1PreparationPRP211

2Server maintenanceSRM111

3Hardware maintenance Lab #01HM14-2

4Software maintenance Lab #01SM1-53





9SetupSTP2

21

4- Base Calendar (Working periods)

SaturdaySundayMondayTuesdayWednesdayThursdayFriday

XXXXXX 1/01/04

Holidays:20 to 21 Jan. 2004

Required:

1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?

3. Construct the corresponding smoothed worker loading?4. Construct the corresponding worker leveling?

5. Construct the target action plan?.

6. Construct the cost profile & S-curve?7. Construct the target master plan?Materials Requirements Planning (MRP) for Maintenance

Case (11): A monthly maintenance plan for 50 similar equipment to replace the gear box for these equipment. The gear box structure is shown below.

ComponentABCDEFG

Lead time (week)1211232

On-Hand101520101050

Required:

1. Time-phased for the gear box structure

2. Gross requirements plan for 50 gear box

3. Net material requirements plan for 50 gear box.Case (12): The monthly plan and the actual maintenance spare parts in ABC Company are as follows:

Spare part #Plan (Jan. 2001)Actual (Jan. 2001)

Planned quantity

(unit)Standard cost (L.E./unit)Actual quantity

(unit)Actual cost

(L.E./unit)

A11401000401100

A12301200201200

A1350900401000

A142085010800

A152095020900

Based on these data, determine the different performance indicators.

Total Maintenance Control

Case (13):Monthly production information on Foundry Shop FS510 was as follows:

ItemJan.

2004Feb.

2004

Working days3128

Standard production rate (ton/hr)88

Average daily time (hr/day)2424

Average down time (hr/day)64

Average standby (hr/day)33

Average target quantity (ton/day)120136

Average actual quantity (ton/day)80105

Average sound quantity (ton/day)7098

Average defect quantity (ton/day)107

Average energy consumption

(1000 kwh/day)4967

Material cost (1000 L.E/day)100130

Based on these data, determine the different PE indicators for the productive system.

Basic data

ItemJan 04Feb 04Feb. / Jan.

Production rate (ton/hr)88100 %

Total time (hr/day)2424100 %

Average down time (hr/day)6467 %

Average available time (hr)1820111 %

Average standby (hr/day)33100 %

Average used time (hr/day)1517113 %

Average target quantity (ton/day)120136113 %

Average actual quantity (ton/day)80105125 %

Average sound quantity (ton/day)7098129 %

Average defect quantity (ton/day)10 (14%)7

(7%)64 %

Energy productivity (kwh/ton)70068498 %

Material productivity (1000 L.E/ton)1429132692 %

Performance Evaluation

IndicatorJanuary

2004February

2004Feb. / Jan.

Availability

18/24= 75 %20/24= 83 %111 %

Performance

efficiency80/120= 67 %105/136= 77 %115 %

Quality rate

70/80= 88 %98/105= 93 %106 %

Utilization ratio

15/18= 83 %17/20= 85 %102 %

Uptime (hr/day)

70/8= 8.7598/8= 12.25140 %

Uptime ratio

8.75/15= 49%12.25/17=72 %147 %

OEE

44 %60 %136 %

TEEP

37 %51 %138 %

NEE

29 %52 %179 %

Energy productivity (kwh/ton)70068498 %

Material productivity (1000 L.E/ton)1429132692 %

Case (14):

The six-monthly maintenance costs ($1000) for a productive system are as follows:

Target Costs:

Cost itemMonth #

JanFebMarAprMayJunJly

PM Cost:

Spar parts

Labor100

50100

50100

50100

50100

50100

50100

50

CM Cost:

Spar parts

Labor200

150200

150200

150200

150200

150200

150200

150

DT Cost300300300300300300300

Actual Costs:

Cost itemMonth #


PM Cost:

Spar parts

Labor23

3238

6549

9656

9468

9465

9054

72

CM Cost:

Spar parts

Labor231

503213

370181

293185

164199

201196

193157

142

DT Cost407397320290330320362

Based on these data, determine the different performance evaluation indicators for the maintenance system.

Target:

Cost itemMonth #

JanFebMarAprMayJunJlyTotal

PM Cost1501501501501501501501050

CM Cost3503503503503503503502450

TM Cost8008008008008008008005600

DT Cost3003003003003003003002100

TM+DT11001100110011001100110011007700

PM/TM0.140.140.140.140.140.140.140.955

CM/PM2.332.332.332.332.332.332.3316.33

Actual:

Cost itemMonth #


PM Cost55103145150162155126896

CM Cost7345834743494003692993208

TM Cost119610839397898928647876550

DT Cost4073973202903303203622426

TM+DT16031480125910791222118411498976

PM/TM0.050.100.150.190.180.180.161.007

CM/PM13.355.663.272.332.472.382.3731.82

Change %:

Cost itemMonth #


PM Cost

CM Cost

TM Cost

DT Cost

TM+DT

PM/TM

CM/PM

Case (15):The yearly PM programs information for six similar gas turbines in a power station are as follows:

Target work performed:

ItemPMCMTotal

Total labor force (worker)18725

Annual spare parts cost ($1000)26472336

Annual labor cost ($1000)----75

Overhead cost ($1000)----514

Average down time

(day/year per gas turbine)511566

Average downtime cost rate = $ 1000 per day

Actual work performed:

ItemPMCMTotal

Total labor force (worker)201030

Annual spare parts cost ($1000)300100400

Annual labor cost ($1000)----80

Overhead cost ($1000)----520

Average down time

(day/year per gas turbine)45550


Performance Evaluation Sheet:

ItemTargetActualChange %

Total labor force (worker)2530+ 20

Annual s. parts cost ($1000)336400+ 19

Annual labor cost ($1000)7580+ 6.6

Overhead cost ($1000)514520+ 1.2

Total m. cost ($1000)9251000+ 8.1

Average down time6650- 24.3

Down time cost ($1000)6650- 24.3

TMC + DTC9911050+ 6.0

Availability %81.986.3+ 5.3

CM/PM % (labor force)7/18 =

38.910/20 =

50+ 28.5

CM/PM % (Spare parts)72/264 =

27.3100/300 =

33.3+ 22.0

Overhead %514/411=

1.25520/480=

1.08- 13.6

Labor productivity %

(worker/gas turbine)25/6=

4.1730/6=

5.00- 16.6

Case (16):The six-monthly maintenance costs ($1000) for a productive system are as follows:

Target Costs:

Cost itemMonth #


PM Cost:

Spar parts

Labor100

50100

50100

50100

50100

50100

50100

50

CM Cost:

Spar parts

Labor200

150200

150200

150200

150200

150200

150200

150

DT Cost300300300300300300300

Actual Costs:

Cost itemMonth #


PM Cost:

Spar parts

Labor23

3238

6549

9656

9468

9465

9054

72

CM Cost:

Spar parts

Labor231

503213

370181

293185

164199

201196

193157

142

DT Cost407397320290330320362


Target:

Cost itemMonth #


PM Cost150150150150150150150

CM Cost350350350350350350350

DT Cost300300300300300300300

TM Cost800800800800800800800

Actual:

Cost itemMonth #


PM Cost 55 103145150162155126

CM Cost 734 583474349400369299

DT Cost 407 397320290330320362

TM Cost11961083939789892864787

7- Machine Failure Analysis

Parameters used for detection of machine faults

Type of faultParameters

VibrationTemp.Oil

Out of balancexxx--

Misalignment / bent shaftxxxx-

Damage of rolling bearingxxxxxx

Damage of journal bearingxxxxxx

Damage of gear boxxxxxxx

Belt problemsxx--

Motor problemsxxx-

Mechanical loosenessxxxxx

Resonancexxx--

xxx

High, easy and soft to measure.

xx

Medium to measure.

x

Low to measure.

-

Non.

Parameters used for detection of pump faults

Type of faultParameters

VibrationTemp.Oil

Out of balancexxx--

Misalignment / bent shaftxxxx-

Damage of rolling bearingxxxxxx

Damage of journal bearingxxxxxx

Damage of gear boxxxxxxx

Belt problemsxx--

Motor problemsxxx-

Mechanical loosenessxxxxx

Resonancexxx--

Minimum flow / Cavitationsxxxxx-

xxx

High, easy and soft to measure.

xx

Medium to measure.

x

Low to measure.

-

Non.

Bearing Failure Analysis

Bearing Failure: Causes and Cures

Excessive Loads:

Excessive loads usually cause premature fatigue. Tight fits, brinelling and improper preloading can also bring about early fatigue failure.

The solution is to reduce the load or redesign using a bearing with greater capacity.

Overheating: Symptoms are discoloration of the rings, balls, and cages from gold to blue.

Temperature in excess of 400F can anneal the ring and ball materials.

The resulting loss in hardness reduces the bearing capacity causing early failure.

In extreme cases, balls and rings will deform. The temperature rise can also degrade or destroy lubricant.

True Brinelling:

Brinelling occurs when loads exceed the elastic limit of the ring material.

Brinell marks show as indentations in the raceways which increase bearing vibration (noise).

Any static overload or severe impact can cause brinelling.

False Brinelling:

False brinelling - elliptical wear marks in an axial direction at each ball position with a bright finish and sharp demarcation, often surrounded by a ring of brown debris indicates excessive external vibration.

Correct by isolating bearings from external vibration, and using greases containing antiwear additives.

Normal Fatigue Failure:

Fatigue failure - usually referred to as spalling - is a fracture of the running surfaces and subsequent removal of small discrete particles of material.

Spalling can occur on the inner ring, outer ring, or balls.

This type of failure is progressive and once initiated will spread as a result of further operation. It will always be accompanied by a marked increase in vibration.

The remedy is to replace the bearing or consider redesigning to use a bearing having a greater calculated fatigue life.

Reverse Loading:

Angular contact bearings are designed to accept an axial load in one direction only.

When loaded in the opposite direction, the elliptical contact area on the outer ring is truncated by the low shoulder on that side of the outer ring.

The result is excessive stress and an increase in temperature, followed by increased vibration and early failure.

Corrective action is to simply install the bearing correctly.

Contamination:

Contamination is one of the leading causes of bearing failure.

Contamination symptoms are denting of the bearing raceways and balls resulting in high vibration and wear.

Clean work areas, tools, fixtures, and hands help reduce contamination failures.

Keep grinding operations away from bearing assembly areas and keep bearings in their original packaging until you are ready to install them.

Lubricant Failure:

Discolored (blue/brown) ball tracks and balls are symptoms of lubricant failure. Excessive wear of balls, ring, and cages will follow, resulting in overheating and subsequent catastrophic failure.

Ball bearings depend on the continuous presence of a very thin -millionths of an inch - film of lubricant between balls and races, and between the cage, bearing rings, and balls.

Failures are typically caused by restricted lubricant flow or excessive temperatures that degrade the lubricants properties.

Corrosion:

Red/brown areas on balls, race-way, cages, or bands of ball bearings are symptoms of corrosion.

This condition results from exposing bearings to corrosive fluids or a corrosive atmosphere.

In extreme cases, corrosion can initiate early fatigue failures.

Correct by diverting corrosive fluids away from bearing areas and use integrally sealed bearings whenever possible.

Misalignment:

Misalignment can be detected on the raceway of the nonrotating ring by a ball wear path that is not parallel to the raceways edges.

If misalignment exceeds 0.001 in./in you can expect an abnormal temperature rise in the bearing and/or housing and heavy wear in the cage ball-pockets.

Appropriate corrective action includes: inspecting shafts and housings for runout of shoulders and bearing seats; use of single point-turned or ground threads on non hardened shafts and ground threads only on hardened shafts; and using precision grade locknuts.

Loose Fits:

Loose fits can cause relative motion between mating parts. If the relative motion between mating parts is slight but continuous, fretting occurs.

Fretting is the generation of fine metal particles which oxidize, leaving a distinctive brown color. This material is abrasive and will aggravate the looseness. If the looseness is enough to allow considerable movement of the inner or outer ring, the mounting surfaces (bore, outer diameters, faces) will wear and heat, causing noise and runout problems.

Tight Fits:

A heavy ball wear path in the bottom of the raceway around the entire circumference of the inner ring and outer ring indicates a tight fit.

Where interference fits exceed the radial clearance at operating temperature, the balls will become excessively loaded. This will result in a rapid temperature rise accompanied by high torque.

Continued operation can lead to rapid wear and fatigue.

Corrective action includes a decrease in total interference.

Case (17): Pump Failure Analysis

Pump Station: PS01

8 Centrifugal pump

Failure Type: Bearing failure

Part code: xxxxx

PM every 1600 R.H. (change oil , filter and bearing)

Bearing failures for centrifugal pumps

(Year 2004)

# of failureEquipment codeRun time

(hr)Repair time (hr)Failure

Mechanism

1100712508Corrosion

2100814506Corrosion

31001100010Temperature

4100415007Corrosion

5100610004Oil

6100212507Corrosion

710037009Oil



10100612509Corrosion

111001100010Oil





1610037006Oil


18100110008Oil

Based on these data, Determine the different PE indicators for this system. Construct how to analyze and eliminate the bearing failure.Failure Analysis:

Pump Station: 8 Centrifugat pumpCode: 1000

Failure Type: Bearing failure

Part code: xxxxx

(Year 2004)

# of failureEquipment codeRun time

(hr)Repair time (hr)Failure

Mechanism

1100712508Corrosion

2100814506Corrosion


4100415007Corrosion

5100610004Oil

6100212507Corrosion

710037009Oil




111001100010Oil





1610037006Oil


18100110008Oil

Total18200145

MTBF = 18200/18 = 1011 hr ( = 0.989 * 10-5 failure/hr

MTTR =145 /18 = 8 hr

A =1011/(1011+8) =99.21%

MTBF at which less than 20 % of the pumps are assumed to fail

Run time

hrFrequencyCumulative FrequencyC.F.

%

1500115.56

14502316.67

12504738.89

100051266.67

70031583.33

60021794.44

500118100

100066.67

?

80.00

MTBF = 760 hr

70083.33

Max. running time =1650 hr. Min. running time= 300 hr

Run time

hrFrequencyMid pointC.F.C.F.

%

1650-1400 31525316.67

1400-115041275738.89

1150-900510251266.67

900-65037751583.33

650-300347518100

Freq

6

5

4

3

2

1

300

650650

900900

11501150

14001400

1650

MTBF

Equipment Level:

Equipment codeMTBF

(hr)MTTR (hr)A

%Failure

Mechanism

10011000

1000

1000

100010

10

8

9.3399.00Temperature

Oil

Oil

10021250

1450

13507

8

7.599.44Corrosion

Corrosion

1003700

700

7009

6

7.598.94Oil

Oil

10041500

1250

13257

11

999.33Corrosion

Corrosion

1005700

1000

8508

9

8.599.01Temperature

Corrosion

10061000

1250

11254

9

6.599.43Oil

Corrosion

10071250

600

9258

8

899.14Corrosion

Temperature

10081450

500

600

8506

8

9

7.6699.10Corrosion

Temperature

Temperature

Average1011899.21Corrosion 8

Oil 5

Temperature 5

Failure Mechanism Level:

Failure

MechanismMTBF

(hr)MTTR (hr)RangesEquipment code

Corrosion1500

1450

1450

1250

1250

1250

1250

10007

8

6

8

7

9

11

9MTBF

1000 -

1500

MTTR

6 111004

1002

1008

1007

1002

1006

1004

1005

Oil1000

1000

1000

700

7004

10

8

9

6MTBF

700 -

1000

MTTR

4-101006

1001

1001

1003

1003

Temperature1000

700

600

600

50010

8

8

9

8MTBF

500 -

1000

MTTR

8-101001

1005

1007

1008

1008

Average1011899.21Corrosion 8

Oil 5

Temperature 5

Remedy:

Maintenance Policy

Condition BasedTime Based

Every 300 hours

Oil analysis

Temperature analysis

Vibration analysis(1) Change oil every 600 hour

(2) Change bearing & oil every

1200 hour

Down time: (1) 1 hr & (2) 8 hr

Cost Analysis:

Cost elements:

Spare parts cost = 1000 L.E./failure

PM impact = 2000 L.E./failure CM impact = 4000 L.E./failure

ParameterCurrentProposed

PM frequency (failure/year)-18

CM frequency (failure/year)181

Spare parts cost (1000 L.E. / year)1819

PM impact (1000 L.E. / year)-36

CM impact (1000 L.E. / year)724

PM & CM impact (1000 L.E. / year)7240

Total cost (1000 L.E. / year)9059

Cost ratio %10065.5

Cost saving %-34.5

Maintenance Policy:

I- Vibration analysis:

1- Frequency: Every 300 Running Hours

2- Tool:

Vibration Equipment: accelerometers, charge amplifier and analyser.

Computer program for trend analysis and prediction.3- International Standard: CDA/MS/NVSH107

4- Method:

1. Record the vibration spectrum, specify the peaks corresponds to the bearing components

2. Record each component peak and frequency.

3. By using the soft ware and the standard limits, determine the trend of each peak.

4. Determine the bearing state(good need service need change)

5- Limits: According to CDA/MS/NVSH107

1. Pre-failure: vibration level5.6 m/s

2. Failure: vibration level 5.610 m/s

3. Near catastrophic failure: vibration level >10 m/s

6- Actions:

1. Bearing is Good

2. Call for bearing change

3. Bearing must be changed immediately

II- Temperature analysis:


2- Tool:

Temperature measuring equipments as thermocouple or infrared camera.

Computer program for trend analysis and prediction.3- International Standard: SKF

4- Method:

Measure the temperature of the bearing on line and take the average value every day.

By using the software analyze the data, determine the max. & average temperature values.

According to the allowable range specified in SKF standard, determine the bearing state.

5- Limits: According to CDA/MS/NVSH10

1. Pre-failure: ( 100 C)

2. Faiulre: (100 125 C)3. Near catastrophic failure: (>125 C)

6- Actions:

1. Bearing is good

2. Call for bearing change

3. Bearing must be changed immediately

III- Oil analysis:


Viscosity change

Acidic content

Wear rate

2- Tool:

Viscometer, PH meter, and particle counter

Computer program for trend analysis and prediction.3- International Standard: ASTM-445 & 664 & 3984- Method:

Take a suitable oil sample volume, to be used in analyses After each 300 hours.

Put it in a closed container, isolated from air, heat and contamination exposes.

Measure the previous mentioned properties then enter the obtained data to the software to be trended.

5- Limits:Viscosity: according to ASTMD-445:

1. Pre-failure: Viscosity change

00 fundamentals maintenance management 2006

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